Question

In an isobaric process where $\gamma$ is the adiabatic exponent of the gas,

• A. The heat given to the gas is $\frac{7}{\gamma -1}$ times the work done by the gas.
• B. The work done by the gas is $(\gamma -1)$ times the change in intermal energy.
• C. The temperature of the gas is increased.
• D. The temperature of the gas is decreased.

Answer 1

Answer: (A), (B), (C)

The correct options are
A The heat given to the gas is $\frac{7}{\gamma -1}$ times the work done by the gas.
B The work done by the gas is $(\gamma -1)$ times the change in intermal energy.
C The temperature of the gas is increased.

At constant pressure, $V\propto T$
If volume increases, temperature also increases.

In an isobaric process, $\Delta Q=\gamma \Delta U$
($\because C_P=\gamma C_V$)
$\Delta Q=\Delta U+\Delta W$ (1st Law)
$\therefore \Delta W=\gamma \Delta U-\Delta U=(\gamma -1)\Delta U$
$\Delta U=\left(\frac{1}{\gamma -1}\right)\Delta W$
$\therefore \Delta Q=\frac{\gamma }{(\gamma -1)}\Delta W$